Let $Y$ be a random variable such that $E[Y] = \lambda$, $\lambda \in \mathbb{R}$ and $E[Y^2]<\infty$. The problem is to find a lower bound on the probability $$ P \left[|Y| > \frac{|\lambda|}{2} \right]. $$
Any leads would be appreciated.
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$\begingroup$ Do you have knowledge on the $\mathbb{E}[|Y|]$ and $\mathbb{E}[|Y|^{2}]$, because in that case the Paley-Zygmund inequality might be what you want $\endgroup$– Fiodor1234Jul 27, 2021 at 9:37
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2 Answers
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Edit : This answer applies to the original question, that was : "for ANY random variable Y, what is the lower bound of (formula)"
The lower bound is 0.
Let's take $(Y_{n})$ a series following a Bernoulli law such as : $P(Y_{n} = 1) =1/n$. Then, ${E(Y_{n})= \lambda = 1/n}$.
$P(\lvert Y_{n} \rvert > \frac{\lvert \lambda \rvert}{2}) = P(\lvert Y_{n} \rvert > \frac{1}{2n}) = P(Y_{n} =1) = \frac{1}{n} < \frac{2}{n}\underset{n\to +\infty}{\longrightarrow} 0 $
Then, $0 < P(\lvert Y_{n} \rvert > \frac{\lvert \lambda \rvert}{2}) < \frac{2}{n}$ , so asymptotically, the lower bound is 0.
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$\begingroup$ How does a series of different random variables respond to a question about a single random variable?? $\endgroup$– whuber ♦Jul 27, 2021 at 18:47
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1$\begingroup$ @whuber my answer was posted before the edit of galen. The question was : "let Y ANY random variable". Then, the question is to find a lower bound for all the possible random variables, and my answer with series provides the response : the lower bound is 0 $\endgroup$– AdrienJul 27, 2021 at 20:52
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$\begingroup$ You still misinterpret what "any" means: sure, the RV can be arbitrary: but a sequence of different random variables with different values of $\lambda$ is not "any" random variable! As such, your analysis sheds no light on the question. If, instead, you were to scale your examples suitably, you would exhibit variables with a common value of $\lambda$ and as a set (not a series) they would demonstrate the nonexistence of any positive lower bound for a given $\lambda.$ $\endgroup$– whuber ♦Jul 28, 2021 at 15:26
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Given $\lambda,$ there must be a universal lower bound $p(\lambda)$ (because $0$ will certainly work.) The question is whether there are any $\lambda$ where this bound exceeds $0.$ Regardless, being a lower bound means that for any random variable $Y$ with $E[Y^2]\lt \infty$ and $E[Y]=\lambda,$ $$\Pr(|Y| \gt |\lambda|/2) \ge p(\lambda).\tag{*}$$
Consider $Y=\lambda X/q$ where $X$ is a Bernoulli$(q)$ variable and $\lambda \ne 0.$ Because $E[X]=q,$ $E[Y] = \lambda (q)/q = \lambda.$ $Y$ also has finite second moment because it is bounded. Moreover, since $q \lt 2,$ the event "$|Y|\gt|\lambda|/2$" equals the event "$Y=\lambda/q.$" We may therefore conclude from $(*)$ that
$$q = \Pr(X=1) \ge \Pr(Y=\lambda/q) = \Pr(|Y| \gt |\lambda|/2) \ge p(\lambda).$$
If we choose $q = p(\lambda)/2,$ this says
$$p(\lambda)/2 \ge p(\lambda),$$
which is true only for non-positive numbers $p(\lambda).$ (You can check that this example still works when $\lambda=0,$ for then $Y$ reduces to the atom at $0$ and $\Pr(|Y|\gt |\lambda|/2) = \Pr(Y \ne 0) = 0.$) Therefore $0$ is the only universal lower bound, no matter what the value of $\lambda$ might be.
